3.1.82 \(\int \frac {(c+d x^3)^2}{(a+b x^3)^{2/3}} \, dx\) [82]

Optimal. Leaf size=132 \[ \frac {2 d (2 b c-a d) x \sqrt [3]{a+b x^3}}{5 b^2}+\frac {d x \sqrt [3]{a+b x^3} \left (c+d x^3\right )}{5 b}+\frac {\left (5 b^2 c^2-5 a b c d+2 a^2 d^2\right ) x \left (1+\frac {b x^3}{a}\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{5 b^2 \left (a+b x^3\right )^{2/3}} \]

[Out]

2/5*d*(-a*d+2*b*c)*x*(b*x^3+a)^(1/3)/b^2+1/5*d*x*(b*x^3+a)^(1/3)*(d*x^3+c)/b+1/5*(2*a^2*d^2-5*a*b*c*d+5*b^2*c^
2)*x*(1+b*x^3/a)^(2/3)*hypergeom([1/3, 2/3],[4/3],-b*x^3/a)/b^2/(b*x^3+a)^(2/3)

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Rubi [A]
time = 0.05, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {427, 396, 252, 251} \begin {gather*} \frac {x \left (\frac {b x^3}{a}+1\right )^{2/3} \left (2 a^2 d^2-5 a b c d+5 b^2 c^2\right ) \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{5 b^2 \left (a+b x^3\right )^{2/3}}+\frac {2 d x \sqrt [3]{a+b x^3} (2 b c-a d)}{5 b^2}+\frac {d x \sqrt [3]{a+b x^3} \left (c+d x^3\right )}{5 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x^3)^2/(a + b*x^3)^(2/3),x]

[Out]

(2*d*(2*b*c - a*d)*x*(a + b*x^3)^(1/3))/(5*b^2) + (d*x*(a + b*x^3)^(1/3)*(c + d*x^3))/(5*b) + ((5*b^2*c^2 - 5*
a*b*c*d + 2*a^2*d^2)*x*(1 + (b*x^3)/a)^(2/3)*Hypergeometric2F1[1/3, 2/3, 4/3, -((b*x^3)/a)])/(5*b^2*(a + b*x^3
)^(2/3))

Rule 251

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, (-b)*(x^n/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

Rule 252

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^Fra
cPart[p]), Int[(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 396

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*x*((a + b*x^n)^(p + 1)/(b*(n*(
p + 1) + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 427

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[d*x*(a + b*x^n)^(p + 1)*((c
+ d*x^n)^(q - 1)/(b*(n*(p + q) + 1))), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, x]

Rubi steps

\begin {align*} \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{2/3}} \, dx &=\frac {d x \sqrt [3]{a+b x^3} \left (c+d x^3\right )}{5 b}+\frac {\int \frac {c (5 b c-a d)+4 d (2 b c-a d) x^3}{\left (a+b x^3\right )^{2/3}} \, dx}{5 b}\\ &=\frac {2 d (2 b c-a d) x \sqrt [3]{a+b x^3}}{5 b^2}+\frac {d x \sqrt [3]{a+b x^3} \left (c+d x^3\right )}{5 b}-\frac {(4 a d (2 b c-a d)-2 b c (5 b c-a d)) \int \frac {1}{\left (a+b x^3\right )^{2/3}} \, dx}{10 b^2}\\ &=\frac {2 d (2 b c-a d) x \sqrt [3]{a+b x^3}}{5 b^2}+\frac {d x \sqrt [3]{a+b x^3} \left (c+d x^3\right )}{5 b}-\frac {\left ((4 a d (2 b c-a d)-2 b c (5 b c-a d)) \left (1+\frac {b x^3}{a}\right )^{2/3}\right ) \int \frac {1}{\left (1+\frac {b x^3}{a}\right )^{2/3}} \, dx}{10 b^2 \left (a+b x^3\right )^{2/3}}\\ &=\frac {2 d (2 b c-a d) x \sqrt [3]{a+b x^3}}{5 b^2}+\frac {d x \sqrt [3]{a+b x^3} \left (c+d x^3\right )}{5 b}+\frac {\left (5 b^2 c^2-5 a b c d+2 a^2 d^2\right ) x \left (1+\frac {b x^3}{a}\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{5 b^2 \left (a+b x^3\right )^{2/3}}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(304\) vs. \(2(132)=264\).
time = 12.95, size = 304, normalized size = 2.30 \begin {gather*} -\frac {x \left (1+\frac {b x^3}{a}\right )^{2/3} \Gamma \left (\frac {4}{3}\right ) \left (-3920 a c^2 \Gamma \left (\frac {1}{3}\right )-1960 a c d x^3 \Gamma \left (\frac {1}{3}\right )-560 a d^2 x^6 \Gamma \left (\frac {1}{3}\right )+3780 a c^2 \Gamma \left (\frac {10}{3}\right )+1890 a c d x^3 \Gamma \left (\frac {10}{3}\right )+540 a d^2 x^6 \Gamma \left (\frac {10}{3}\right )-270 a \left (14 c^2+7 c d x^3+2 d^2 x^6\right ) \Gamma \left (\frac {10}{3}\right ) \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {10}{3};-\frac {b x^3}{a}\right )+297 b c^2 x^3 \Gamma \left (\frac {10}{3}\right ) \, _2F_1\left (\frac {4}{3},\frac {5}{3};\frac {13}{3};-\frac {b x^3}{a}\right )+432 b c d x^6 \Gamma \left (\frac {10}{3}\right ) \, _2F_1\left (\frac {4}{3},\frac {5}{3};\frac {13}{3};-\frac {b x^3}{a}\right )+135 b d^2 x^9 \Gamma \left (\frac {10}{3}\right ) \, _2F_1\left (\frac {4}{3},\frac {5}{3};\frac {13}{3};-\frac {b x^3}{a}\right )+81 b x^3 \left (c+d x^3\right )^2 \Gamma \left (\frac {10}{3}\right ) \, _3F_2\left (\frac {4}{3},\frac {5}{3},2;1,\frac {13}{3};-\frac {b x^3}{a}\right )\right )}{1260 a \left (a+b x^3\right )^{2/3} \Gamma \left (\frac {1}{3}\right ) \Gamma \left (\frac {10}{3}\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(c + d*x^3)^2/(a + b*x^3)^(2/3),x]

[Out]

-1/1260*(x*(1 + (b*x^3)/a)^(2/3)*Gamma[4/3]*(-3920*a*c^2*Gamma[1/3] - 1960*a*c*d*x^3*Gamma[1/3] - 560*a*d^2*x^
6*Gamma[1/3] + 3780*a*c^2*Gamma[10/3] + 1890*a*c*d*x^3*Gamma[10/3] + 540*a*d^2*x^6*Gamma[10/3] - 270*a*(14*c^2
+ 7*c*d*x^3 + 2*d^2*x^6)*Gamma[10/3]*Hypergeometric2F1[1/3, 2/3, 10/3, -((b*x^3)/a)] + 297*b*c^2*x^3*Gamma[10
/3]*Hypergeometric2F1[4/3, 5/3, 13/3, -((b*x^3)/a)] + 432*b*c*d*x^6*Gamma[10/3]*Hypergeometric2F1[4/3, 5/3, 13
/3, -((b*x^3)/a)] + 135*b*d^2*x^9*Gamma[10/3]*Hypergeometric2F1[4/3, 5/3, 13/3, -((b*x^3)/a)] + 81*b*x^3*(c +
d*x^3)^2*Gamma[10/3]*HypergeometricPFQ[{4/3, 5/3, 2}, {1, 13/3}, -((b*x^3)/a)]))/(a*(a + b*x^3)^(2/3)*Gamma[1/
3]*Gamma[10/3])

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (d \,x^{3}+c \right )^{2}}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^3+c)^2/(b*x^3+a)^(2/3),x)

[Out]

int((d*x^3+c)^2/(b*x^3+a)^(2/3),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+c)^2/(b*x^3+a)^(2/3),x, algorithm="maxima")

[Out]

integrate((d*x^3 + c)^2/(b*x^3 + a)^(2/3), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+c)^2/(b*x^3+a)^(2/3),x, algorithm="fricas")

[Out]

integral((d^2*x^6 + 2*c*d*x^3 + c^2)/(b*x^3 + a)^(2/3), x)

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Sympy [C] Result contains complex when optimal does not.
time = 1.98, size = 126, normalized size = 0.95 \begin {gather*} \frac {c^{2} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {2}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}} \Gamma \left (\frac {4}{3}\right )} + \frac {2 c d x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}} \Gamma \left (\frac {7}{3}\right )} + \frac {d^{2} x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}} \Gamma \left (\frac {10}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**3+c)**2/(b*x**3+a)**(2/3),x)

[Out]

c**2*x*gamma(1/3)*hyper((1/3, 2/3), (4/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(2/3)*gamma(4/3)) + 2*c*d*x**4*gam
ma(4/3)*hyper((2/3, 4/3), (7/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(2/3)*gamma(7/3)) + d**2*x**7*gamma(7/3)*hyp
er((2/3, 7/3), (10/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(2/3)*gamma(10/3))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+c)^2/(b*x^3+a)^(2/3),x, algorithm="giac")

[Out]

integrate((d*x^3 + c)^2/(b*x^3 + a)^(2/3), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (d\,x^3+c\right )}^2}{{\left (b\,x^3+a\right )}^{2/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x^3)^2/(a + b*x^3)^(2/3),x)

[Out]

int((c + d*x^3)^2/(a + b*x^3)^(2/3), x)

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